Problem: Let $A = (1,1)$ be a point on the parabola $y = x^2.$  The normal to the parabola at $A$ is drawn, intersecting the parabola again at $B.$  Find $B.$

[asy]
unitsize(1 cm);

pair A, B;

A = (1,1);
B = (-3/2,9/4);

real parab (real x) {
  return(x^2);
}

draw(graph(parab,-2,2));
draw((A + (-1,-2))--(A + (1,2)));
draw((A + (1,-0.5))--(A + (-3,1.5)));
draw(rightanglemark(A + (-3,1.5), A, A + (1,2), 10));

dot("$A$", A, S);
dot("$B$", B, SW);
[/asy]

Note: The normal at a point $P$ on a curve $\mathcal{C}$ is the line passing through $P$ that is perpendicular to the tangent to $\mathcal{C}$ at $P.$
Then the equation of the tangent at $A = (1,1)$ is of the form
\[y - 1 = m(x - 1),\]or $y = mx - m + 1.$  Substituting into $y = x^2,$ we get
\[mx - m + 1 = x^2.\]Then $x^2 - mx + m - 1 = 0.$  Since we have a tangent, this quadratic should have a double root.  And since the $x$-coordinate of $A$ is $1,$ the double root is $x = 1.$  Hence, this quadratic is identical to $(x - 1)^2 = x^2 - 2x + 1,$ which means $m = 2.$

Then the slope of the normal is $-\frac{1}{2},$ so the equation of the normal is
\[y - 1 = -\frac{1}{2} (x - 1).\]We want the intersection of the normal with $y = x^2,$ so we set $y = x^2$:
\[x^2 - 1 = -\frac{1}{2} (x - 1).\]We can factor the left-hand side:
\[(x - 1)(x + 1) = -\frac{1}{2} (x - 1).\]The solution $x = 1$ corresponds to the point $A.$  Otherwise, $x \neq 1,$ so we can divide both sides by $x - 1$:
\[x + 1 = -\frac{1}{2}.\]Hence, $x = -\frac{3}{2},$ so $B = \boxed{\left( -\frac{3}{2}, \frac{9}{4} \right)}.$